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SPOJ Number of Palindromes(回文树)

Time Limit: 100MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

Description

Each palindrome can be always created from the other palindromes, if a single character is also a palindrome. For example, the string "malayalam" can be created by some ways:

* malayalam = m + ala + y + ala + m
* malayalam = m + a + l + aya + l + a + m

We want to take the value of function NumPal(s) which is the number of different palindromes that can be created using the string S by the above method. If the same palindrome occurs more than once then all of them should be counted separately.

Input

The string S.

Output

The value of function NumPal(s).

Limitations

0 < |s| <= 1000

Example

Input:

malayalam

Output:

Hint

Added by:	The quick brown fox jumps over the lazy dog
Date:	2010-10-18
Time limit:	0.100s-0.170s
Source limit:	50000B
Memory limit:	1536MB
Cluster:	Cube (Intel G860)
Languages:	All
Resource:	Udit Agarwal

回文树：

参考大牛的博客：http://blog.csdn.net/lwfcgz/article/details/48739051

回文树是一种处理回文串的强大工具

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
#define MAX 1005
struct Node
{
    int next[26];
    int len;
    int sufflink;
    int num;
}tree[MAX];
char s[MAX];

int num;
int suff;
bool addLetter(int pos)
{
    int cur=suff,curlen=0;
    int let=s[pos]-'a';

    while(1)
    {
        curlen=tree[cur].len;
        if(pos-1-curlen>=0&&s[pos-1-curlen]==s[pos])
            break;
        cur=tree[cur].sufflink;
    }
    if(tree[cur].next[let])
    {
        suff=tree[cur].next[let];
        return false;
    }
    num++;
    suff=num;
    tree[num].len=tree[cur].len+2;
    tree[cur].next[let]=num;
    if(tree[num].len==1)
    {
        tree[num].sufflink=2;
        tree[num].num=1;
        return true;
    }
    while(1)
    {
        cur=tree[cur].sufflink;
        curlen=tree[cur].len;
        if(pos-1-curlen>=0&&s[pos-1-curlen]==s[pos])
        {
            tree[num].sufflink=tree[cur].next[let];
            break;
        }
    }
    tree[num].num=1+tree[tree[num].sufflink].num;
    return true;

}
void initTree()
{
    num=2;suff=2;
    tree[1].len=-1;tree[1].sufflink=1;
    tree[2].len=0;tree[2].sufflink=1;
}
int main()
{
    scanf("%s",s);
    int len=strlen(s);
    initTree();
    long long int ans=0;
    for(int i=0;i<len;i++)
    {
         addLetter(i);
         ans+=tree[suff].num;
    }
    printf("%d
",ans);
    return 0;
}

Number of Palindromes